Why symplectic geometry is the natural setting for classical mechanics

Of course everything in this essay is common knowledge among symplectic geometers (as well as many other people). I'm writing it down here because when I first learned it I didn't understand it intuitively, so I thought it would be amusing to try to give as simple-minded an explanation as I can.

My goal is to give a precise formulation of the idea of the phase space of a system in classical mechanics. This space is supposed to be a manifold with some additional structure. The points correspond to states of the system, and the additional structure is essentially the laws of motion. The dynamics of the system will be determined by a function on the manifold called the Hamiltonian (it gives the energy of each state of the system).

The fundamental example to keep in mind is n particles in three-dimensional space, acted on by some force depending on their relative positions. The phase space is (R6)n, since each particle is specified by six coordinates: three position coordinates and three momentum coordinates. The additional structure is supposed to correspond to Newton's laws (this idea will be made more precise below). The dynamics are determined from the potential function for the force, i.e. the Hamiltonian.

In more general settings there is no reason why the underlying manifold should be Euclidean space. For example, particles may be constrained to lie on curves or surfaces, or the states may not even be described in terms of a bunch of particles. It's natural to ask how abstractly this can be set up. Can we axiomatize the properties of phase space, and then just deal with it abstractly (without worrying about the precise setup that led to it)? In fact we can, and symplectic manifolds are the appropriate abstraction. In what follows I'll attempt to explain where symplectic manifolds come from. Whenever I write "should," I am indicating an intuitive argument.

Suppose M is a smooth manifold. What additional structure is needed to define dynamics on M? We want to turn a Hamiltonian H (i.e., any smooth function on M) into a vector field V. Then the dynamics consists of the flow along integral curves of V.

First, note that the vector field V should depend only on the differential dH. This makes perfect sense, since the dynamics should depend only on how the energy is changing locally, and not for example on a global additive constant.

Second, V should depend linearly on dH. This amounts to the assertion that Newton's laws are linear differential equations, which is true.

What sort of object turns a 1-form dH into a vector field V in this way? The most natural such object is a tensor field, in particular a section of Hom(T*M,TM). Here TM denotes the tangent bundle of M and T*M denotes the cotangent bundle. It will actually be more convenient to use the dual approach, namely Hom(TM,T*M), which is of course the same as T*M tensor T*M. It works as follows:

Let ω be a section of T*M tensor T*M. Over each point of M, ω is a bilinear mapping from pairs to vectors to the field R of real numbers. The 1-form dH gives us a covector at each point, and the vector field V should be such that plugging it into ω gives dH. More precisely, for each vector field X, ω(V,X) = (dH)(X).

Our first requirement is that ω must be nondegenerate, so we can always solve for V. Specifically, we require that for every 1-form ν on M, there is a vector field V such that ω(V,X) = ν(X) for all vector fields X. (It follows from linear algebra that V is always unique if it always exists.)

So far we've got a manifold M with a nondegenerate tensor field ω in T*M tensor T*M. However, not every such tensor field will do, since we really haven't used much information about the laws of physics yet. The next restriction is conservation of energy. Under which conditions on ω does conservation of energy hold?

What it means in our situation is that the Hamiltonian H is constant along the flow lines of our vector field V. In other words, (dH)(V) = 0. In terms of ω, that means ω(V,V)=0. This must hold for all vector fields V coming from Hamiltonians, which means that the bilinear form above each point of M must be an alternating form. (Not every vector field comes from a Hamiltonian, but each tangent vector occurs in some Hamiltonian vector field.) In other words, ω is a differential 2-form on M.

A symplectic manifold is a manifold together with a closed, nondegenerate 2-form. We have seen that the 2-form corresponds to having laws of physics given by linear differential equations that conserve energy, and nondegeneracy is needed to ensure that the equations have solutions. All that remains is to explain why ω should be closed, i.e., why dω = 0. Unfortunately I don't see how to explain that without slightly more notation, but it won't be too bad.

This requirement corresponds to a slightly more subtle issue, namely that the laws of physics should not depend on time. Let Ft denote the time-t flow along the vector field corresponding to some Hamiltonian H. One natural way to get the laws of physics at time t is to look at the pullback Ft*ω. We want it to equal ω. Clearly it does so at time 0, so to check that it always does we simply differentiate with respect to t, as follows.

If V is the vector field corresponding to H, then the derivative of Ft*ω with respect to t is Ft*LVω, where LV denotes the Lie derivative with respect to V. Let iV denote the interior product with V, so for example dH = iVω. Then LVω = iV(dω) + d(iVω) = iV(dω) + d(dH) = iV(dω) (because d2 = 0). The requirement that ω be closed means that LVω = 0. Then Ft*ω does not depend on t, so it always equals ω, as desired.

I've swept one tiny issue under the rug. One might wonder why this setup seems to be using first-order differential equations when Newton's laws are typically stated as second-order equations. That just amounts to the usual trick of replacing a second-order differential equation in terms of position with a pair of first-order equations in terms of position and velocity (here, momentum).